Question

The actual width of a 2×4 piece of lumber is approximately 1 3/4 in (1.75 in). but can vary considerably. Larry’s Lumber advertises consistent dimensions for better building, and claims all 2×4 sold have a mean width of 1 3/4 in. with a standard deviation of 0.02 in.
a. [5 marks] Assume the distribution of widths is approximately normal. Find a symmetric interval about the mean that contains almost all of the 2×4 widths. Show your work. (1.75
b. [5 marks] Suppose a random 2×4 has width 1.78 in. Is there any evidence to suggest the Larry’s Lumber claim is wrong? Justify your answer.
c. [5 marks] Suppose a random 2×4 has width 1.68 in. Is there any evidence to suggest the Larry’s Lumber claim is wrong? Justify your answer.

Answer 1

Answer:

Given,

mean = 1.75

standard deviation = 0.02

a)

CI = ( - 3 , + 3)

substitute values

= (1.75 - 3*0.02 , 1.75 + 3*0.02)

= (1.69 , 1.81)b)

b)

Null hypothesis Ho : u = 1.78

Alternative hypothesis Ha : u != 1.78

Here at 5% level of significance, z value is +/- 1.96

consider,

z = (x - ) /

substitute values

= (1.75 - 1.78) / 0.02

z = - 1.5

Here we can say that, the claim is correct.

c)

Null hypothesis Ho : u = 1.68

Alternative hypothesis Ha : u != 1.68

Here at 5% level of significance, z value is +/- 1.96

consider,

z = (x - ) /

substitute values

= (1.75 - 1.68) / 0.02

= 3.5

Here we observe that, z > 1.96, so we can say that the claim is not right.